Experiment 4

Work-Heat Conversion Factor

(The Joule Heat)

 

Objective:

 

The objective of this experiment is to verify that 1 calorie = 4.186 joules.

 

Equipment:

 

A calorimeter with electric heating element, a dc power source, two multi-meters, a thermometer, a few connecting wires, water, a stop watch, and a mass scale

 

Theory:

 

In the past several decades, it has been repeatedly verified by many scientists and experimenters that when 1 calorie of heat is converted to mechanical or electric work, 4.18J of work is generated and vice versa.

 

1cal = 4.18J.

 

The heating element in a calorimeter may be connected to a power source for a certain period during which electric work (energy) causes an increase in the temperature of the fluid in the calorimeter. The result is the conversion of electric work to heat energy.  To calculate the energy delivered to the calorimeter, the formula for power may be used:

 

P = W/t        or        W = P t .

 

Since electric power is P = V I, the above equation that gives the electric energy consumption becomes:

 

W = V I t .       (1)

 

On one hand, by measuring the applied voltage V, the current I, and the consumption time t, the electric energy delivered to the calorimeter can be calculated from this formula.  On the other hand, by measuring the initial and final temperatures of the calorimeter, as well as the masses of the water, the aluminum container, and the heating element (including its attachments), the amount of heat generated can be calculated.  Of course, the specific heats of water, aluminum, as well as the heating element are also needed in heat calculation.  The heat absorbed in the calorimeter is

 

Qtotal = Qwater + QAl + Qelement .

 

Note that the heating element itself has a small mass, and the mass of it may be added to the mass of the metal legs to which it is attached.  This is not going to introduce significant error.  Therefore, it is better to measure the mass of the heating element assembly together, and simply use the specific heat of brass for it, although the element itself  may have a different specific heat.  The existing heating element assembly in the calorimeter in this laboratory is mainly brass.  The above equation may therefore be written as:

 

Qtotal = Qwater + QAl + Qbrass .

or

Qtotal  = mw cw ΔT  +  mAl cAl ΔT  +  mbr cbr ΔT.

 

Qtotal= [ mw cw + mAl cAl + mbr cbr ] ΔT.    (2)

 

This equation calculates the total heat generated in calories as a result of electric work done in the heating element in joules.  Both values can be calculated in order to determine the # of joules one calorie can generate or vice versa.

 

Procedure:

 

Using a zeroed mass scale, measure the mass of the internal container of the calorimeter mAl first, and then measure it again when it is 1/4 - 1/3 filled with clean water.  This way, the mass of added water mw can be calculated easily.  Disassemble the heating element assembly from the lid of the calorimeter and measure its mass mbr with a balance scale.  Reassemble the heating element into the lid.  If the mass is predetermined, skip this step and consult your instructor.  Place a thermometer in the calorimeter through the hole in the lid, and wait until thermal equilibrium is reached.  Record the initial temperature Ti .  With the electric power source off, connect the calorimeter to the source in series with an ammeter. 

 

Note that the ammeter in this experiment is to measure a current that exceeds 200mA.  There is a third terminal on the multi-meter that is specially used for this purpose.  The positive wire (red) must be pulled out of the usual positive terminal and inserted into the third terminal that is labeled 10A.  The negative wire (black) remains in the terminal that is labeled "Common."

 

Now that you have connected the calorimeter, the ammeter (in +10A position), and the power source in series, it is almost time to turn the power on.  The voltmeter on the power source is not accurate.  It is better to measure the voltage of the power source with a separate meter.  A typical diagram of the experimental arrangement is shown below:

 

 

Now disconnect the ammeter from the power source for the purpose of first selecting an operating voltage.  Turn on the power source, and set the voltage to not more than 6V Turn off the power (without changing the voltage setting), and reconnect the ammeter to the power source.  Re-check the initial temperature Ti .  When you are ready to measure time, turn the power source on.

 

It is better to write down the values of V and I every 30 seconds, and then calculate a mean value for each, and use those mean values in the calculation of consumed electric energy W = V I t.  This will compensate (to some extent) for voltage fluctuations of the power source due to causes out of your control.  The following table may be used for this purpose:

 

t (sec) 10 20 30 40 50 60 .... .... .... ....
I (Amps)                    
V(volts)                    

 

It is a good idea to keep shaking the calorimeter gently (or stirring the water), once every minute to facilitate heat distribution.  After the desired period of heating is reached, turn off the power source and wait for the temperature to reach its maximum.  Record this maximum as Tf  .

 

Data:

Given:            

            cw = 1.000     cal /(gr oC )

            cBr = 0.0924  cal /(gr oC)

            cAl = 0.215    cal /(gr oC)

            t = 10.00 minutes as an example.

(W/Q)accepted = 4.18 J/cal

 

             Measured:   mw , mbr ,  mAl , Ti ,  Tf  , V , I, and t

 

 

Calculations:

Use equations (1) and (2) to calculate W and Q.  Then find (W/Q)measured .

Comparison of the Results:

The accepted and measured values of (W/Q) may be used to obtain a percent error.

 

Conclusion: To be explained by students.

Discussion:   To be explained by students.